Homework 14.5 part d

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MadisonFuentes1G
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Joined: Fri Sep 29, 2017 7:04 am

Homework 14.5 part d

Postby MadisonFuentes1G » Thu Feb 21, 2019 12:34 pm

For letter d) in problem 14.5 (6th edition), the answe key says to multiply the first half-reaction by 3, but then the new reaction contains 4P4, not 3P4.

Is this a mistake in the answer key or the actual answer? If it’s correct, could someone explain why?

Lauren Ho 2E
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Joined: Tue Oct 09, 2018 12:16 am

Re: Homework 14.5 part d

Postby Lauren Ho 2E » Thu Feb 21, 2019 2:30 pm

The Solution Manual Errors PDF on Professor Lavelle's website doesn't point to any errors in the solution manual for problem 14.5, so I would assume that the given answer is right

MadisonFuentes1G
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Joined: Fri Sep 29, 2017 7:04 am

Re: Homework 14.5 part d

Postby MadisonFuentes1G » Fri Feb 22, 2019 12:44 pm

I checked the Errors PDF as well, but I still have no idea how they got 4P4 by multiplying by 3.

204929947
Posts: 76
Joined: Fri Apr 06, 2018 11:03 am

Re: Homework 14.5 part d

Postby 204929947 » Fri Feb 22, 2019 10:08 pm

Also, in part b, how come when they balance the equation out, they get rid of Br2 and only write 6OH- --> 5Br-(aq)+BrO3-(aq)+3H2O(l) ???
how did they cancel out Br2??

Matthew Tran 1H
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Joined: Fri Sep 28, 2018 12:16 am

Re: Homework 14.5 part d

Postby Matthew Tran 1H » Fri Feb 22, 2019 11:39 pm

MadisonFuentes1G wrote:For letter d) in problem 14.5 (6th edition), the answe key says to multiply the first half-reaction by 3, but then the new reaction contains 4P4, not 3P4.

Is this a mistake in the answer key or the actual answer? If it’s correct, could someone explain why?


The overall redox reaction has 4P4 because you multiply the oxidation half-reaction by 3, giving 3P4, and add it with the reduction half-reaction that has 1P4. Therefore in total there is 4P4. I have the 7th edition so I'm assuming the solution manual answer hasn't changed from 6th to 7th. Hope this helps!


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